Problem: The value of $\sqrt{68}$ lies between which two consecutive integers ? Integers that appear in order when counting, for example 2 and 3.
Explanation: Consider the perfect squares near $68$ . [ What are perfect squares? Perfect squares are integers which can be obtained by squaring an integer. The first 13 perfect squares are: $ 1,4,9,16,25,36,49,64,81,100,121,144,169$ $64$ is the nearest perfect square less than $68$ $81$ is the nearest perfect square more than $68$ So, we know $64 < 68 < 81$ So, $\sqrt{64} < \sqrt{68} < \sqrt{81}$ So $\sqrt{68}$ is between $8$ and $9$.